Помогите решить 164,спасибо

Решите задачу:

$a)x - \frac{1}{4} = 7 \frac{1}{3} - 3 \frac{1}{12} = \\ = 7 \frac{4}{12} - 3 \frac{1}{12} = 4 \frac{3}{12} = 4 \frac{1}{4}$

$x = 4 \frac{1}{4} + \frac{1}{4} \\ x = 4 \frac{2}{4} = 2 \frac{1}{2} = 2.5$

$b)x + \frac{1}{4} = 6 \frac{2}{3} - 2 \frac{1}{6} = \\ 6 \frac{4}{6} - 2 \frac{1}{6} = 4 \frac{3}{6} = 4 \frac{1}{2}$

$x = 4 \frac{1}{2} - \frac{1}{4} = 4 \frac{2}{4} - \frac{1}{4} \\ x = 4 \frac{1}{4}$